WAEC Physics 2016 Theory — Question 8

Model answer

(a) The net force is the sum of all forces acting on an object. It is the vector sum of all forces which are acting on any object. When two or more forces are acting on a particle, the net force comes into account. When the forces acting on an object are balanced, the net force is zero and the object will move with uniform velocity, i.e. zero acceleration. (b) The principle of conservation of linear momentum states that in any system of colliding objects, the total momentum of the system before collision is equal to the total momentum of the system after collision, provided there is no net external force acting on the system. Examples: (i) Collision of two bodies (ii) firing of a gun (c) 200g=0.2kg Velocity at which ball hits ground: v²=u²+2gs = 0+2×10×2 → v=√40=6.32 ms⁻¹ (downward) Velocity of rebound: v²=u²-2gs (rising 1.8m) → v=√(2×10×1.8)=√36=6 ms⁻¹ (upward, opposite direction) Impulse = m(vf-vi) = 0.2×(-6-6.32) = 0.2×(-12.32) = -2.464 kg m/s The impulse received by the floor is 2.464 kg·m/s (the negative sign shows the change in momentum is directed upward, i.e. opposite to the initial motion). (d)(i) v=ω√(A²-x²); v=200cm/s=2m/s, x=10cm=0.1m ω=2×3.14×5=31.4 rad/s⁻¹ 2=31.4√(A²-0.1²) A²-0.1²=(2/31.4)²=0.00405≈0.0046 A²=0.01+0.0046=0.0146 A=√0.0146=0.121m (ii) Maximum velocity, from V=ω√(A²-x²), Vmax occurs when x=0: Vmax=ωA=31.4×0.121=3.8 m/s (iii) Maximum potential energy = Total energy = ½kA² = ½ω²mA² (since w²=k/m, k=w²m) = ½×(31.4)²×0.02×(0.121)² = 0.142 J